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3 years ago

3ln(2x) + 7 = 15. how do I do this

Mathematics

1 answer:

andrey2020 [161]3 years ago

5 0

Answer:

Step-by-step explanation:

3 ln(2x) + 7 = 15 Subtract 7 from both sides

3 ln(2x) = 15 - 7 Combine

3 ln(2x) = 8 Divide by 3

ln(2x) = 8/3 Do the division. Take the antilog of both sides.

e^ln(2x) = e^(8/3)

2x = 14.392 Divide by 2

x = 14.392

x = 7. 196

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What value of x makes the equation 3 = <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B3x%2B%206%7D" id="TexFormula1" title="

disa [49]

3^3=3x+6,so3x=27-6=21,sox=7

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2 years ago

Read 2 more answers

How to differentiate ?

Bas_tet [7]

Use the power, product, and chain rules:

$y = x^2 (3x-1)^3$

• product rule

$\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}$

• power rule for the first term, and power/chain rules for the second term:

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}$

• power rule

$\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3$

Now simplify.

$\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}$

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

$\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) = \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)$

Differentiate both sides and you end up with the same derivative:

$\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2$

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2 years ago

Which are true or false

Savatey [412]

we are given

$y=19.75x$

Here,

y is the number of gallons of water in the pool

x is the number of minutes the truck has been filing the pool

now, we will verify each options

option-A:

When x=1

then y=19.75

so, for 1 minute , number gallons =19.75

so, this is FALSE

option-B:

Since, we have

$y=19.75x$

It is similar to equation of line

y=mx+b

But b=0

It means that it passes through origin

so, this is TRUE

option-C:

$y=19.75x$

we can compare it with

y=mx

where m is the rate

so, we get

$m=19.75$

while we are given

the water in swimming pool increase about 100 gallons by every 5 minutes

so, we get rate as

$m=\frac{100}{5} =20$ gallons per minute

this is close to 19.75

so, they are approximately equal

so, this is TRUE

option-d:

x-value and y-value will show relationship

and ratio will always be constant or equivalent

because it has linear relationship

so, this is TRUE

option-e:

We need to check

(8,150) ..so, x=8 and y=150

we can plug this in our equation

and we get

$150=19.75\times 8$

$150=158$

Since, left side is not equal to right side

so, this can not be point

so, this is FALSE

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2 years ago

Which equation represents a line that is perpendicular to the y-axis?

Sindrei [870]

Answer:

C, $y=-8$

Step-by-step explanation:

$y=-8$ , is a horizontal line, which makes it vertical to the y-axis

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2 years ago

A certain forest covers an area of 4800 km^2 . Suppose that each year this area decreases by 5.25% . What will the area be after

9966 [12]

Answer:

the area after 6 years is 3,473 km^2

Step-by-step explanation:

The computation of the area after 6 years is as follows:

= Area × (1 - decreased percentage)^number of years

= 4,800 km^2 × (1 - 5.25%)^6

= 4,800 km^2 × 0.9475^6

= 3,473 km^2

Hence, the area after 6 years is 3,473 km^2

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2 years ago

3ln(2x) + 7 = 15. how do I do this​

3ln(2x) + 7 = 15. how do I do this